mysql - 查找一个或多个用户之间的对话

Question

我在查找给定 user_id 之间的对话时遇到问题。

SQL 表:

+-------------------+
| conversation_user |
+-------------------+
| conversation_id   |
| user_id           |
+-------------------+

我试过了

SELECT `conversation_id` FROM `conversation_user` WHERE `user_id` IN (X, Y) HAVING COUNT(*) = N

但它不能正常工作。知道如何选择正确的 conversation_id 吗？对话可以在一个或多个用户之间进行。

编辑:

+-----------------+---------+
| conversation_id | user_id |
+-----------------+---------+
|               1 |       1 | 
|               1 |       2 | 
+-----------------+---------+
|               2 |       1 | 
|               2 |       3 | 
+-----------------+---------+
|               3 |       1 | 
+-----------------+---------+
|               4 |       1 | 
|               4 |       2 | 
|               4 |       3 | 
+-----------------+---------+

假设我想获取用户 1 和 2 之间的对话。结果必须是 1，而不是 1 and 4 或 4。

Answer 1

我觉得你缺GROUP BY子句

<罢工>

<罢工>

SELECT `conversation_id` 
FROM `conversation_user` 
WHERE `user_id` IN (X, Y) 
GROUP BY conversation_id
HAVING COUNT(DISTINCT user_id) = N

<罢工>

或

SELECT `conversation_id` 
FROM `conversation_user` a
WHERE `user_id` IN (X, Y) 
GROUP BY conversation_id
HAVING COUNT(DISTINCT user_id) = 
    (
        SELECT COUNT(DISTINCT userid)
        FROM `conversation_user` b
        WHERE b.`conversation_id` = a.`conversation_id`
        GROUP BY b.`conversation_id`
    )

SQLFiddle Demo